Limits and Derivatives... 唔識呀~

(1) lim(x→-1) [(x^3)+1]/(x+1)
lim(x→-1) [(x+1)(x^2 –x + 1)]/(x+1)
lim(x→-1) (x^2 –x + 1)
lim(x→-1) ((-1)^2 –(-1) + 1)
lim(x→-1) (1 +1 + 1) = 3

(2) lim(x→1) [(x^3)-(x^2)+x-1]/[(x^2)+x-2]
lim(x→1) [(x^2(x-1)+1(x-1) ]/[(x + 2)(x – 1)]
lim(x→1) [(x^2+1)(x-1) ]/[(x + 2)(x – 1)]
lim(x→1) (x^2+1)/(x + 2)
lim(x→1) (1^2+1)/(1 + 2) = 2/3

公式!? : f ' (a) = lim(h→0) [f(a+h)-f(a)]/h ←唔明

 I don’t have enough space to explain the concept to you here. Look it up in any Calculus textbook or ask your teacher to clarify it. 

For each function f(x), find the differential coefficient for each value of x

(1) f(x) = (x-2)^2 (x=-1)
f(x) = x^2 – 4x + 4
f ’(x) = 2x – 4
f ‘ (-1) = 2(-1) – 4
f ‘ (-1) = -6

(2) f(x) = (x^3)-2x+4 (x=2)

Given f(x)=x^2, find the following differential coefficient

Since f ‘(x) = 2x (i.e. dy/dx = 2x)
(1) f ' (3) = 2(3) = 6

(2) f ' (0) = 2(0) = 0

(3) f ' (-1) = 2(-1) = -2

The derivative of f(x)
f ' (x) = lim (h→0) [f(x+h)-f(x)]/h

Obtain the derivative of each function using the above formula
(1) f(x) = 2x^2
lim (h→0) [2(x+h)^2-2x^2]/h
lim (h→0) [2(x^2 +2xh +h^2) - 2x^2]/h
lim (h→0) [2x^2 +4xh +2h^2 - 2x^2]/h
lim (h→0) [4xh +2h^2 ]/h
lim (h→0) 4x +2h
lim (h→0) 4x +2(0) = 4x

(2) f(x) = (x^2)+1 (Don’t have space) The answer is 2x

(3) f(x) = (x^2)+2x (Don’t have space) The answer is 2x + 2

(4) f(x) = 2x
lim (h→0) [2(x+h) -2x]/h
lim (h→0) [2x + 2h -2x]/h
lim (h→0) [2h]/h = 2

(5) f(x)=x
lim (h→0) [(x+h) -x]/h
lim (h→0) [x + h -x]/h
lim (h→0) [h]/h = 1

(6) f(x)=x^3
lim (h→0) [(x+h)^3-x^3]/h
lim (h→0) [(x^2 +2xh +h^2)(x+h) - x^3]/h
lim (h→0) [x^3 + 3x^2h +3xh^2 +h^3 - x^3]/h
lim (h→0) [3x^2h +3xh^2 +h^3]/h
lim (h→0) [3x^2 +3xh +h^2]
lim (h→0) 3x^2 +3x(0) +(0)^2 = 3x^2

(7) f(x)=x^4
lim (h→0) [(x+h)^4-x^4]/h
lim (h→0) [(x+h)^2(x+h)^2 -x^4]/h
lim (h→0) [(x^2 +2xh +h^2) (x^2 +2xh +h^2) -x^4]/h
lim (h→0) [(x^4 +4x^3h +6x^2h^2 +4xh^3+ h^4) -x^4]/h
lim (h→0) [4x^3h +6x^2h^2 +4xh^3 + h^4]/h
lim (h→0) 4x^3 +6x^2h +4xh^2 + h^3
lim (h→0) 4x^3 +6x^2(0) +4(0)h^2 +(0)^3 = 4x^3